Question about tensor products, decomposable tensors, ... I need some help with the following problem:

Let $V_1,\ldots,V_m$ be finite dimensional vector spaces over $\mathbb{K}$.
Let $\varphi \in L(V_1,\ldots,V_m;U)$ such that $Im(\varphi)=U$.
Show that there exists a subspace $K$ of $V_1 \otimes \cdots \otimes V_m$ such that every class of $\frac{V_1 \otimes \cdots \otimes V_m}{K}$ contains a decomposable tensor.

I have some ideas but none of them work, such as:
$$U = Im(\varphi) \simeq \frac{V_1 \times \cdots \times V_m}{Ker(\varphi)}$$
which is wrong because $\varphi$ is multilinear and not linear, so I have no idea how to solve this problem.
Any ideas? Hints? Thanks!

If you know where I can find this problem (maybe in some book) I would be very grateful.

 A: By the universal property of the tensor product, $\varphi$ induces a linear map $\psi\colon V_1 \otimes \cdots \otimes V_m \to U$ which is also surjective (why?). And in fact we know more: for any $u \in U$, because $\varphi$ is surjective there are $v_i \in V_i$ such that $\varphi(v_1, \dots, v_m) = \psi(v_1 \otimes \cdots \otimes v_m) = u$. I'm hoping that from here you can fill in a gap. Thinking about the first isomorphism theorem is a good idea, though I don't think you need to apply it directly.
A: By the universal factorization property of $\otimes$, there exists a unique linear map $\psi: V_1 \otimes \cdots \otimes V_m \longrightarrow U$ such that $\varphi = \psi \circ \otimes$.
As $\varphi$ is surjective, $\psi$ is surjective: given $u \in U$, $u = \varphi(v_1,\ldots,v_m)$ for some $(v_1,\ldots,v_m) \in V_1 \times \cdots \times V_m$, then $u = \psi \circ \otimes (v_1,\ldots,v_m) = \psi(v_1 \otimes \cdots \otimes v_m)$, then $u \in Im(\psi)$.
So given $u \in U$, there exists $(v_1,\ldots,v_m) \in V_1 \times \cdots \times V_m$ such that
\begin{align*}
u = \psi(v_1 \otimes \cdots \otimes v_m) = \varphi(v_1,\ldots,v_m).
\end{align*}
Let $K=Ker(\psi)$.
Then,
\begin{align*}
\pi: \frac{V_1 \otimes \cdots \otimes V_m}{K} &\to U\\
z + Ker(\psi) &\mapsto \psi(z)
\end{align*}
is an isomorphism (easy to check).
Let us prove that each class of $\frac{V_1 \otimes \cdots \otimes V_m}{K}$ contains a decomposable tensor.
If $z + K = K$ it is trivial because $0 \in K$ and $0$ is decomposable.
Suppose that there exists a class $z_0 + K \neq K$ such that $z_0 + K$ does not contain a decomposable tensor.
Then, it does not exist $z \in z_0 + K$ decomposable such that $\pi(z + K) = \psi(z) = u$, but this is false, since each $u \in U$ is of the form $\psi(v_1 \otimes \cdots \otimes v_m)$ for some $(v_1,\ldots,v_m) \in V_1 \times \cdots \times V_m$ as we saw above.
So we conclude that each class of $\frac{V_1 \otimes \cdots \otimes V_m}{K}$ contains a decomposable tensor.
